In fig- QT - PR - QR - QS and - 1- 2 - Prove that PQS - TQR-

∠ 1 = ∠ 2 0ex(Given) ⇒ PR = PQ ex ...(i) (Sides opposite to equal angles in QRP) Also QT PR = QR QS (Given) ...(ii) From (i) and (ii), we have QT PR=QR QS⇒QP Q ...

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In fig, QT / ...

Question

In fig, $\frac{Q T}{P R} = \frac{Q R}{Q S}$ and $∠ 1 = ∠ 2$ . Prove that $△ P Q S \sim △ T Q R$ .

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Solution

$∠ 1 = ∠ 2$ (Given)
$\Rightarrow$ PR = PQ $. . . (i)$
(Sides opposite to equal angles in $△ Q R P$ )
Also $\frac{Q T}{P R} = \frac{Q R}{Q S}$ (Given) ...(ii)

From (i) and (ii), we have
$\frac{Q T}{P R} = \frac{Q R}{Q S} \Rightarrow \frac{Q P}{Q T} = \frac{Q S}{Q R}$ ...(iii)
Now, in triangle PQR and TQR $(e a c h = ∠ 1)$
and $\frac{P Q}{T Q} = \frac{Q S}{Q R} (f r o m (3))$
$\Rightarrow △ P Q S \sim △ T Q R (S A S S i m i l a r i t y)$

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In fig, QTPR=QRQS and ∠1=∠2. Prove that △PQS∼△TQR.

∠1=∠2(Given) ⇒ PR = PQ ...(i) (Sides opposite to equal angles in △QRP) Also QTPR=QRQS (Given) ...(ii) From (i) and (ii), we have QTPR=QRQS⇒QPQT=QSQR ...(iii) Now, in triangle PQR and TQR (each=∠1) and PQTQ=QSQR(from(3)) ⇒△PQS∼△TQR(SASSimilarity)

In fig, $\frac{Q T}{P R} = \frac{Q R}{Q S}$ and $∠ 1 = ∠ 2$ . Prove that $△ P Q S \sim △ T Q R$ .